I Limits of multivariable functions (uniform convergence)

1. Aug 13, 2016

JulienB

Hi everybody! I'm preparing an exam of "Analysis II" (that's how the subject's called in German), and I have trouble understanding how to find the limit of a multivariable function, especially when it comes to proving the uniform convergence. Here is an example given in the script of my teacher:

$f(x,y) = \frac{xy}{x^2 + y^2}$ (1)

He claims that $\lim\limits_{x \to 0} ( \lim\limits_{y \to 0} f(x,y)) = \lim\limits_{y \to 0} ( \lim\limits_{x \to 0} f(x,y))$ but that $\lim\limits_{(x,y) \to (0,0)} f(x,y)$ still doesn't exist without any explanation. Just before he makes the following claims:

Sufficient conditions for $\lim\limits_{(x,y) \to (0,0)} f(x,y)$ not to exist:
(i) The limits $\lim\limits_{x \to 0} ( \lim\limits_{y \to 0} f(x,y))$ and $\lim\limits_{y \to 0} ( \lim\limits_{x \to 0} f(x,y))$ both exist but are unequal.
(ii) There exists sequences $x_n \to x_0$ and $y_n \to y_0$ so that the sequence $f(x_1, y_1), f(x_2, y_2),...$ does not converge.
(iii) The exists sequences $x_n \to x_0$, $\bar{x}_n \to x_0$, $y_n \to y_0$ and $\bar{y}_n \to y_0$ so that the sequences $f(x_1, y_1), f(x_2, y_2),...$ and $f(\bar{x}_1, \bar{y}_1), f(\bar{x}_2, \bar{y}_2),...$ converge but have different limits.

One sufficient condition for $\lim\limits_{(x,y) \to (0,0)} f(x,y)$ to exist:
$\lim\limits_{x \to 0} ( \lim\limits_{y \to 0} f(x,y))$ exists and the functions $f(x,\cdot)$ converge uniformly towards the function $g$ on the interval $(0,r)$. The function $g$ is defined by $\lim\limits_{x \to 0} f(x,y) = g(y)\ \forall y \in (0,r)$.

The definition of uniform convergence is given at a different place as:

$\forall \epsilon > 0\ \exists \delta > 0\ \forall y \in (0,r)\ \forall x \in (0, \delta): |f(x,y) - g(y)| \leq \epsilon$.

Can you help me make sense of this? As I understand it, a uniformly converging sequence $f_n$ is a sequence for which the objects tend towards the same function $f$ at the same speed (independently from $x$). Here are my thoughts about the example given above (1):

$\lim\limits_{x \to 0} ( \lim\limits_{y \to 0} f(x,y)) = 0 = \lim\limits_{y \to 0} ( \lim\limits_{x \to 0} f(x,y))$

The limits are indeed equal, and $g(y) = \lim\limits_{x \to 0} f(x,y) = 0\ \forall y \in (0, + \infty)$. Is that correct? Then if I take the definition of uniform convergence, I must show that for all $\epsilon$, there exists a strictly positive $\delta$ so that for all $y \in (0,+ \infty)$ and for all $x \in (0, \delta)$ holds good: $|f(x,y) - g(y)| = |f(x,y)| \leq \epsilon$. Yeah well... I've tried showing the opposite using (ii) but I could not find two sequences that converge so that the sequence of functions diverges. Any tips/clarification about the matter?

Thanks a lot in advance for your answer, I appreciate it!

Julien.

2. Aug 13, 2016

JulienB

I've seen in some Youtube videos people use the "path method". In my previous example I would calculate the limit:

$\lim\limits_{x \to 0} ( \lim\limits_{y \to mx} f(x,y)) = \lim\limits_{x \to 0} \frac{mx^2}{x^2 + m^2 x^2} = \lim\limits_{x \to 0} \frac{m}{1 + m^2} = \frac{m}{1 + m^2} \neq 0\ \forall m \neq 0$

Is that a sufficient proof to show that the limit $\lim\limits_{(x,y) \to (0,0)} f(x,y)$ does not exist?

Thanks a lot for your answers, I'm looking forward.

Julien.

3. Aug 13, 2016

Monci

This is my first non-introduction post in the forums, and I'll be testing how everything works. I hope to be able to answer your question correctly, however.

First of all, your function $f(x) = \frac{xy}{x^2+y^2}$ doesn't converge because it doesn't follow condition (iii). Take, for example, $x_n=y_n=\frac{1}{n}$. Now take $x_n=1/n, y_n=1/n^2$. They give different limits.

The problem with your argument later is that $lim_{x \rightarrow 0} f(x,y) = 0$ is true for every $y$, so there is pointwise convergence, in other words, at every point y, there is a limit. But we want to know that there is uniform convergence. In other words, that for a given $\epsilon$ we can give a $\delta$ which works for every $y \in (0, +\infty)$. Let's show that this can't be done in this case:

Take $\epsilon=1/3$. For any $\delta > 0$, there exists a $y \in (0, +\infty)$ (we take, for example, $y = \delta/2$) such that there exists an $x \in (0, \delta)$ (we take $x = \delta/2$) such that $|{f(x,y)}| > \epsilon$. But $|{f(x,y)}| = 1/2 > \epsilon$.

Hence we have proved the negative of the sufficient condition for existence. This doesn't mean that the function doesn't have a limit at $(0,0)$. It only means that we cannot find it in this way. On the other hand, we have proved a sufficient condition for non-existance, so we are done.

Edit: The intuitive idea behind limits of multivariable functions is that you should be able to approach the solution in any curve you want, and the limit should exist and stay the same. In this case, as you comment, the curve $y = mx$ gives different results depending on $m$, so that is sufficient to prove that there is not a limit here. If this is not clear to you, you should try to reread the definitions of limits of several variables, and think about why it should be true.

Last edited: Aug 13, 2016
4. Aug 13, 2016

JulienB

@Monci Thanks a lot for your answer, that's a brilliant first post! Your proof is very clear to me, but what always ticks me about such proofs is how you found so quickly the sequences. Do you have a systematic method? I searched for almost a hour for such sequences but couldn't find any... I would be miserable in the exam right now. I was convinced though (from the structure of my teacher's script) that this example was related to condition (ii), my mistake I guess.

Also, may I ask what you think about my second post (about the "path method")? Is it legit? This seems to be a very common way of proving a multivariable function does not have a limit, but our teacher did not instruct this method at all. Maybe for us to really understand the theory behind the existence of limits?

Thanks again for your answer, I appreciate it.

Julien.

5. Aug 13, 2016

Monci

I copy my edit in case you didn't see it: The intuitive idea behind limits of multivariable functions is that you should be able to approach the solution in any curve you want, and the limit should exist and stay the same. In this case, as you comment, the curve $y = mx$ gives different results depending on $m$, so that is sufficient to prove that there is not a limit here. If this is not clear to you, you should try to reread the definitions of limits of several variables, and think about why it should be true.

So, a systematic method would be: It is easier to show limit doesn't exist than to show limit exists. Start by trying normal curves. If the limit exists for $y=mx, y=mx^2, y=mx^3$ and is the same, then try to prove the limit always exists. If you cannot prove it exists, try weirder curves, think about the problem in a different way (try to see what techniques people use and practice. A good example of a useful tool would be to go to polar coordinates.) ... In this case, I see that $y=x$ and $y=x^2$ give different limits, so I construct sequences such that:
1 They follow the equation of the curve.
2 They tend to the limit point.
The fact that we can always do this is tied to the definition of "limit point".

I also want to say that your condition (iii) implies (ii). Imagine you have $(x_n), (y_n), (\overline{x_n}), (\overline{y_n})$, with the properties in (iii). Can you construct a sequence with the properties in (ii)? On the other hand, your condition (ii) doesn't imply (iii). Can you give an example where (ii) holds but (iii) doesn't?

6. Aug 13, 2016

JulienB

@Monci thanks again for your explanations. Actually that method is pretty clear to me. If the solution is different for another path, then the limit doesn't exist. And I like your method, it is very clear and understandable. I think more than the theory, it is the lack of clarity in the script of my teacher that got me confused.

It took me a while to answer your question, but after watching an hour of youtube videos about the topic I think I have a better idea of what I'm looking for

We define the sequences $x_n = (-1)^n/n$ and $y_n = 1/n$ which converge towards $x_0 = y_0 = 0$. But the sequence of functions $f(x_n, y_n) = \frac{\frac{(-1)^n}{n^2}}{\frac{(-1)^{2n}}{n^2} + \frac{1}{n^2}} = \frac{(-1)^n}{(-1)^{2n} + 1}$ alternates between $1/2$ and $- \infty$. It is quite far from your original example, but is it valid nevertheless? Hopefully I didn't mess up with the sequences.

Concerning the second question, I am clueless at the moment. But I will give it some thoughts.

Julien.

7. Aug 13, 2016

Monci

$(-1)^{2n}=1$, so in your example the answer alternates between $1/2$ and $-1/2$. The answer is right, however, for this example, but I meant the question in a general manner.

What I meant, is that if you have sequences $(x_n), (y_n), (\overline{x_n}), (\overline{y_n})$, then you can make another two sequences, $(X_n), (Y_n)$, defined as $$X_n = \begin{cases} x_n & \text{ if } n \text{ odd} \\ \overline{x_n} & \text{ otherwise} \end{cases}$$ and $Y_n$ can be defined in a similar manner. Then, we can see that $f(X_n, Y_n)$ has two distinct subsequences that converge to different limits, so it doesn't converge. This works not only in your situation, but in any situation, and states that if condition (iii) holds, then condition (ii) must hold.

To see that (ii) doesn't imply (iii), you'd have to give an example: A limit such that you can prove (ii) but can prove that (iii) isn't satisfied. If you cannot find such an example, ask in this thread and I'll give you one.

8. Aug 13, 2016

mathwonk

it seems natural to choose straight line approaches to the origin, e.g. along an axis, say x=0, and along a diagonal like x=y. try this.

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